There is a quotation below:
For a discrete group $\Gamma$ we let $\lambda:\Gamma\rightarrow B(l^{2}(\Gamma))$ denote the left regular representation: $\lambda_{s}(\delta_{t})=\delta_{st}~$ for all $s, t\in \Gamma$, where $\{\delta_{t}: t\in \Gamma\}\subset l^{2}(\Gamma)$ is the canonical orthonormal basis. There is also a right regular representation $\rho:\Gamma\rightarrow\mathbb{B}(l^{2}(\Gamma))$, defined by $\rho_{s}(\delta_{t})=\delta_{t s^{-1}}$.
My question is what is $\lambda_{s}^{*}$? I guess $\lambda_{s}^{*}=\lambda_{s^{-1}}~$(But I am not sure). And what about $\rho_{s}^{*}$?
Well, since $\{\delta_t\}$ is a orthonormal basis for $\ell^2(\Gamma)$, to find $\lambda_s^\ast$ and $\rho_s^\ast$, it suffices to look at $\langle \lambda_s^\ast \delta_t, \delta_u \rangle$ and $\langle \rho_s^\ast \delta_t, \delta_u \rangle$.
For example, let's find $\lambda_s^\ast$. Now, $$ \langle \lambda_s^\ast \delta_t, \delta_u \rangle := \langle \delta_t, \lambda_s \delta_u \rangle = \langle \delta_t, \delta_{su} \rangle = \begin{cases} 1, &\text{if $t = su$,}\\ 0 ,&\text{else}. \end{cases} $$ But, $t = su$ if and only if $s^{-1}t = u$, so that $$ \langle \lambda_s^\ast \delta_t, \delta_u \rangle = \langle \delta_t, \delta_{su} \rangle = \langle\delta_{s^{-1}t},\delta_u\rangle = \langle \lambda_{s^{-1}}\delta_t,\delta_u\rangle. $$ Hence, $\lambda_s^\ast = \lambda_{s^{-1}}$. Can you play the same game for $\rho_s$?